3.1819 \(\int \frac {(A+B x) (d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=209 \[ -\frac {e^2 (-5 a B e-A b e+6 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} (b d-a e)^{3/2}}-\frac {e \sqrt {d+e x} (-5 a B e-A b e+6 b B d)}{8 b^3 (a+b x) (b d-a e)}-\frac {(d+e x)^{3/2} (-5 a B e-A b e+6 b B d)}{12 b^2 (a+b x)^2 (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

[Out]

-1/12*(-A*b*e-5*B*a*e+6*B*b*d)*(e*x+d)^(3/2)/b^2/(-a*e+b*d)/(b*x+a)^2-1/3*(A*b-B*a)*(e*x+d)^(5/2)/b/(-a*e+b*d)
/(b*x+a)^3-1/8*e^2*(-A*b*e-5*B*a*e+6*B*b*d)*arctanh(b^(1/2)*(e*x+d)^(1/2)/(-a*e+b*d)^(1/2))/b^(7/2)/(-a*e+b*d)
^(3/2)-1/8*e*(-A*b*e-5*B*a*e+6*B*b*d)*(e*x+d)^(1/2)/b^3/(-a*e+b*d)/(b*x+a)

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Rubi [A]  time = 0.18, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {27, 78, 47, 63, 208} \[ -\frac {e^2 (-5 a B e-A b e+6 b B d) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} (b d-a e)^{3/2}}-\frac {(d+e x)^{3/2} (-5 a B e-A b e+6 b B d)}{12 b^2 (a+b x)^2 (b d-a e)}-\frac {e \sqrt {d+e x} (-5 a B e-A b e+6 b B d)}{8 b^3 (a+b x) (b d-a e)}-\frac {(d+e x)^{5/2} (A b-a B)}{3 b (a+b x)^3 (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^(3/2))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

-(e*(6*b*B*d - A*b*e - 5*a*B*e)*Sqrt[d + e*x])/(8*b^3*(b*d - a*e)*(a + b*x)) - ((6*b*B*d - A*b*e - 5*a*B*e)*(d
+ e*x)^(3/2))/(12*b^2*(b*d - a*e)*(a + b*x)^2) - ((A*b - a*B)*(d + e*x)^(5/2))/(3*b*(b*d - a*e)*(a + b*x)^3)
- (e^2*(6*b*B*d - A*b*e - 5*a*B*e)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(8*b^(7/2)*(b*d - a*e)^(3
/2))

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)^{3/2}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac {(A+B x) (d+e x)^{3/2}}{(a+b x)^4} \, dx\\ &=-\frac {(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}+\frac {(6 b B d-A b e-5 a B e) \int \frac {(d+e x)^{3/2}}{(a+b x)^3} \, dx}{6 b (b d-a e)}\\ &=-\frac {(6 b B d-A b e-5 a B e) (d+e x)^{3/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac {(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}+\frac {(e (6 b B d-A b e-5 a B e)) \int \frac {\sqrt {d+e x}}{(a+b x)^2} \, dx}{8 b^2 (b d-a e)}\\ &=-\frac {e (6 b B d-A b e-5 a B e) \sqrt {d+e x}}{8 b^3 (b d-a e) (a+b x)}-\frac {(6 b B d-A b e-5 a B e) (d+e x)^{3/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac {(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}+\frac {\left (e^2 (6 b B d-A b e-5 a B e)\right ) \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{16 b^3 (b d-a e)}\\ &=-\frac {e (6 b B d-A b e-5 a B e) \sqrt {d+e x}}{8 b^3 (b d-a e) (a+b x)}-\frac {(6 b B d-A b e-5 a B e) (d+e x)^{3/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac {(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}+\frac {(e (6 b B d-A b e-5 a B e)) \operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{8 b^3 (b d-a e)}\\ &=-\frac {e (6 b B d-A b e-5 a B e) \sqrt {d+e x}}{8 b^3 (b d-a e) (a+b x)}-\frac {(6 b B d-A b e-5 a B e) (d+e x)^{3/2}}{12 b^2 (b d-a e) (a+b x)^2}-\frac {(A b-a B) (d+e x)^{5/2}}{3 b (b d-a e) (a+b x)^3}-\frac {e^2 (6 b B d-A b e-5 a B e) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{8 b^{7/2} (b d-a e)^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 177, normalized size = 0.85 \[ \frac {\frac {(a+b x) (-5 a B e-A b e+6 b B d) \left (3 \sqrt {b} e^2 (a+b x)^2 \sqrt {d+e x} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )-b (d+e x) \sqrt {a e-b d} (3 a e+2 b d+5 b e x)\right )}{\sqrt {a e-b d}}-8 b^3 (d+e x)^3 (A b-a B)}{24 b^4 (a+b x)^3 \sqrt {d+e x} (b d-a e)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^(3/2))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(-8*b^3*(A*b - a*B)*(d + e*x)^3 + ((6*b*B*d - A*b*e - 5*a*B*e)*(a + b*x)*(-(b*Sqrt[-(b*d) + a*e]*(d + e*x)*(2*
b*d + 3*a*e + 5*b*e*x)) + 3*Sqrt[b]*e^2*(a + b*x)^2*Sqrt[d + e*x]*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-(b*d) +
a*e]]))/Sqrt[-(b*d) + a*e])/(24*b^4*(b*d - a*e)*(a + b*x)^3*Sqrt[d + e*x])

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fricas [B]  time = 1.14, size = 1115, normalized size = 5.33 \[ \left [\frac {3 \, {\left (6 \, B a^{3} b d e^{2} - {\left (5 \, B a^{4} + A a^{3} b\right )} e^{3} + {\left (6 \, B b^{4} d e^{2} - {\left (5 \, B a b^{3} + A b^{4}\right )} e^{3}\right )} x^{3} + 3 \, {\left (6 \, B a b^{3} d e^{2} - {\left (5 \, B a^{2} b^{2} + A a b^{3}\right )} e^{3}\right )} x^{2} + 3 \, {\left (6 \, B a^{2} b^{2} d e^{2} - {\left (5 \, B a^{3} b + A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \sqrt {b^{2} d - a b e} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) - 2 \, {\left (4 \, {\left (B a b^{4} + 2 \, A b^{5}\right )} d^{3} + 2 \, {\left (2 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} d^{2} e - {\left (23 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} d e^{2} + 3 \, {\left (5 \, B a^{4} b + A a^{3} b^{2}\right )} e^{3} + 3 \, {\left (10 \, B b^{5} d^{2} e - {\left (21 \, B a b^{4} - A b^{5}\right )} d e^{2} + {\left (11 \, B a^{2} b^{3} - A a b^{4}\right )} e^{3}\right )} x^{2} + 2 \, {\left (6 \, B b^{5} d^{3} + {\left (5 \, B a b^{4} + 7 \, A b^{5}\right )} d^{2} e - {\left (31 \, B a^{2} b^{3} + 11 \, A a b^{4}\right )} d e^{2} + 4 \, {\left (5 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{48 \, {\left (a^{3} b^{6} d^{2} - 2 \, a^{4} b^{5} d e + a^{5} b^{4} e^{2} + {\left (b^{9} d^{2} - 2 \, a b^{8} d e + a^{2} b^{7} e^{2}\right )} x^{3} + 3 \, {\left (a b^{8} d^{2} - 2 \, a^{2} b^{7} d e + a^{3} b^{6} e^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{7} d^{2} - 2 \, a^{3} b^{6} d e + a^{4} b^{5} e^{2}\right )} x\right )}}, \frac {3 \, {\left (6 \, B a^{3} b d e^{2} - {\left (5 \, B a^{4} + A a^{3} b\right )} e^{3} + {\left (6 \, B b^{4} d e^{2} - {\left (5 \, B a b^{3} + A b^{4}\right )} e^{3}\right )} x^{3} + 3 \, {\left (6 \, B a b^{3} d e^{2} - {\left (5 \, B a^{2} b^{2} + A a b^{3}\right )} e^{3}\right )} x^{2} + 3 \, {\left (6 \, B a^{2} b^{2} d e^{2} - {\left (5 \, B a^{3} b + A a^{2} b^{2}\right )} e^{3}\right )} x\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (4 \, {\left (B a b^{4} + 2 \, A b^{5}\right )} d^{3} + 2 \, {\left (2 \, B a^{2} b^{3} - 5 \, A a b^{4}\right )} d^{2} e - {\left (23 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} d e^{2} + 3 \, {\left (5 \, B a^{4} b + A a^{3} b^{2}\right )} e^{3} + 3 \, {\left (10 \, B b^{5} d^{2} e - {\left (21 \, B a b^{4} - A b^{5}\right )} d e^{2} + {\left (11 \, B a^{2} b^{3} - A a b^{4}\right )} e^{3}\right )} x^{2} + 2 \, {\left (6 \, B b^{5} d^{3} + {\left (5 \, B a b^{4} + 7 \, A b^{5}\right )} d^{2} e - {\left (31 \, B a^{2} b^{3} + 11 \, A a b^{4}\right )} d e^{2} + 4 \, {\left (5 \, B a^{3} b^{2} + A a^{2} b^{3}\right )} e^{3}\right )} x\right )} \sqrt {e x + d}}{24 \, {\left (a^{3} b^{6} d^{2} - 2 \, a^{4} b^{5} d e + a^{5} b^{4} e^{2} + {\left (b^{9} d^{2} - 2 \, a b^{8} d e + a^{2} b^{7} e^{2}\right )} x^{3} + 3 \, {\left (a b^{8} d^{2} - 2 \, a^{2} b^{7} d e + a^{3} b^{6} e^{2}\right )} x^{2} + 3 \, {\left (a^{2} b^{7} d^{2} - 2 \, a^{3} b^{6} d e + a^{4} b^{5} e^{2}\right )} x\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/48*(3*(6*B*a^3*b*d*e^2 - (5*B*a^4 + A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (5*B*a*b^3 + A*b^4)*e^3)*x^3 + 3*(6*B*a
*b^3*d*e^2 - (5*B*a^2*b^2 + A*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (5*B*a^3*b + A*a^2*b^2)*e^3)*x)*sqrt(b^
2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a)) - 2*(4*(B*a*b^4 + 2*A*
b^5)*d^3 + 2*(2*B*a^2*b^3 - 5*A*a*b^4)*d^2*e - (23*B*a^3*b^2 + A*a^2*b^3)*d*e^2 + 3*(5*B*a^4*b + A*a^3*b^2)*e^
3 + 3*(10*B*b^5*d^2*e - (21*B*a*b^4 - A*b^5)*d*e^2 + (11*B*a^2*b^3 - A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 + (5*B
*a*b^4 + 7*A*b^5)*d^2*e - (31*B*a^2*b^3 + 11*A*a*b^4)*d*e^2 + 4*(5*B*a^3*b^2 + A*a^2*b^3)*e^3)*x)*sqrt(e*x + d
))/(a^3*b^6*d^2 - 2*a^4*b^5*d*e + a^5*b^4*e^2 + (b^9*d^2 - 2*a*b^8*d*e + a^2*b^7*e^2)*x^3 + 3*(a*b^8*d^2 - 2*a
^2*b^7*d*e + a^3*b^6*e^2)*x^2 + 3*(a^2*b^7*d^2 - 2*a^3*b^6*d*e + a^4*b^5*e^2)*x), 1/24*(3*(6*B*a^3*b*d*e^2 - (
5*B*a^4 + A*a^3*b)*e^3 + (6*B*b^4*d*e^2 - (5*B*a*b^3 + A*b^4)*e^3)*x^3 + 3*(6*B*a*b^3*d*e^2 - (5*B*a^2*b^2 + A
*a*b^3)*e^3)*x^2 + 3*(6*B*a^2*b^2*d*e^2 - (5*B*a^3*b + A*a^2*b^2)*e^3)*x)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^
2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) - (4*(B*a*b^4 + 2*A*b^5)*d^3 + 2*(2*B*a^2*b^3 - 5*A*a*b^4)*d^2*e - (
23*B*a^3*b^2 + A*a^2*b^3)*d*e^2 + 3*(5*B*a^4*b + A*a^3*b^2)*e^3 + 3*(10*B*b^5*d^2*e - (21*B*a*b^4 - A*b^5)*d*e
^2 + (11*B*a^2*b^3 - A*a*b^4)*e^3)*x^2 + 2*(6*B*b^5*d^3 + (5*B*a*b^4 + 7*A*b^5)*d^2*e - (31*B*a^2*b^3 + 11*A*a
*b^4)*d*e^2 + 4*(5*B*a^3*b^2 + A*a^2*b^3)*e^3)*x)*sqrt(e*x + d))/(a^3*b^6*d^2 - 2*a^4*b^5*d*e + a^5*b^4*e^2 +
(b^9*d^2 - 2*a*b^8*d*e + a^2*b^7*e^2)*x^3 + 3*(a*b^8*d^2 - 2*a^2*b^7*d*e + a^3*b^6*e^2)*x^2 + 3*(a^2*b^7*d^2 -
2*a^3*b^6*d*e + a^4*b^5*e^2)*x)]

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giac [B]  time = 0.23, size = 381, normalized size = 1.82 \[ \frac {{\left (6 \, B b d e^{2} - 5 \, B a e^{3} - A b e^{3}\right )} \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{8 \, {\left (b^{4} d - a b^{3} e\right )} \sqrt {-b^{2} d + a b e}} - \frac {30 \, {\left (x e + d\right )}^{\frac {5}{2}} B b^{3} d e^{2} - 48 \, {\left (x e + d\right )}^{\frac {3}{2}} B b^{3} d^{2} e^{2} + 18 \, \sqrt {x e + d} B b^{3} d^{3} e^{2} - 33 \, {\left (x e + d\right )}^{\frac {5}{2}} B a b^{2} e^{3} + 3 \, {\left (x e + d\right )}^{\frac {5}{2}} A b^{3} e^{3} + 88 \, {\left (x e + d\right )}^{\frac {3}{2}} B a b^{2} d e^{3} + 8 \, {\left (x e + d\right )}^{\frac {3}{2}} A b^{3} d e^{3} - 51 \, \sqrt {x e + d} B a b^{2} d^{2} e^{3} - 3 \, \sqrt {x e + d} A b^{3} d^{2} e^{3} - 40 \, {\left (x e + d\right )}^{\frac {3}{2}} B a^{2} b e^{4} - 8 \, {\left (x e + d\right )}^{\frac {3}{2}} A a b^{2} e^{4} + 48 \, \sqrt {x e + d} B a^{2} b d e^{4} + 6 \, \sqrt {x e + d} A a b^{2} d e^{4} - 15 \, \sqrt {x e + d} B a^{3} e^{5} - 3 \, \sqrt {x e + d} A a^{2} b e^{5}}{24 \, {\left (b^{4} d - a b^{3} e\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

1/8*(6*B*b*d*e^2 - 5*B*a*e^3 - A*b*e^3)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/((b^4*d - a*b^3*e)*sqrt(-
b^2*d + a*b*e)) - 1/24*(30*(x*e + d)^(5/2)*B*b^3*d*e^2 - 48*(x*e + d)^(3/2)*B*b^3*d^2*e^2 + 18*sqrt(x*e + d)*B
*b^3*d^3*e^2 - 33*(x*e + d)^(5/2)*B*a*b^2*e^3 + 3*(x*e + d)^(5/2)*A*b^3*e^3 + 88*(x*e + d)^(3/2)*B*a*b^2*d*e^3
+ 8*(x*e + d)^(3/2)*A*b^3*d*e^3 - 51*sqrt(x*e + d)*B*a*b^2*d^2*e^3 - 3*sqrt(x*e + d)*A*b^3*d^2*e^3 - 40*(x*e
+ d)^(3/2)*B*a^2*b*e^4 - 8*(x*e + d)^(3/2)*A*a*b^2*e^4 + 48*sqrt(x*e + d)*B*a^2*b*d*e^4 + 6*sqrt(x*e + d)*A*a*
b^2*d*e^4 - 15*sqrt(x*e + d)*B*a^3*e^5 - 3*sqrt(x*e + d)*A*a^2*b*e^5)/((b^4*d - a*b^3*e)*((x*e + d)*b - b*d +
a*e)^3)

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maple [B]  time = 0.10, size = 487, normalized size = 2.33 \[ -\frac {\sqrt {e x +d}\, A a \,e^{4}}{8 \left (b e x +a e \right )^{3} b^{2}}+\frac {\sqrt {e x +d}\, A d \,e^{3}}{8 \left (b e x +a e \right )^{3} b}+\frac {\left (e x +d \right )^{\frac {5}{2}} A \,e^{3}}{8 \left (b e x +a e \right )^{3} \left (a e -b d \right )}-\frac {5 \sqrt {e x +d}\, B \,a^{2} e^{4}}{8 \left (b e x +a e \right )^{3} b^{3}}-\frac {11 \left (e x +d \right )^{\frac {5}{2}} B a \,e^{3}}{8 \left (b e x +a e \right )^{3} \left (a e -b d \right ) b}+\frac {11 \sqrt {e x +d}\, B a d \,e^{3}}{8 \left (b e x +a e \right )^{3} b^{2}}-\frac {3 \sqrt {e x +d}\, B \,d^{2} e^{2}}{4 \left (b e x +a e \right )^{3} b}+\frac {5 \left (e x +d \right )^{\frac {5}{2}} B d \,e^{2}}{4 \left (b e x +a e \right )^{3} \left (a e -b d \right )}-\frac {\left (e x +d \right )^{\frac {3}{2}} A \,e^{3}}{3 \left (b e x +a e \right )^{3} b}+\frac {A \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b^{2}}-\frac {5 \left (e x +d \right )^{\frac {3}{2}} B a \,e^{3}}{3 \left (b e x +a e \right )^{3} b^{2}}+\frac {5 B a \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{8 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b^{3}}+\frac {2 \left (e x +d \right )^{\frac {3}{2}} B d \,e^{2}}{\left (b e x +a e \right )^{3} b}-\frac {3 B d \,e^{2} \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{4 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}\, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

1/8*e^3/(b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(5/2)*A-11/8*e^3/(b*e*x+a*e)^3/(a*e-b*d)/b*(e*x+d)^(5/2)*a*B+5/4*e^2/(
b*e*x+a*e)^3/(a*e-b*d)*(e*x+d)^(5/2)*B*d-1/3*e^3/(b*e*x+a*e)^3/b*(e*x+d)^(3/2)*A-5/3*e^3/(b*e*x+a*e)^3/b^2*(e*
x+d)^(3/2)*a*B+2*e^2/(b*e*x+a*e)^3/b*(e*x+d)^(3/2)*B*d-1/8*e^4/(b*e*x+a*e)^3/b^2*(e*x+d)^(1/2)*A*a+1/8*e^3/(b*
e*x+a*e)^3/b*(e*x+d)^(1/2)*A*d-5/8*e^4/(b*e*x+a*e)^3/b^3*(e*x+d)^(1/2)*B*a^2+11/8*e^3/(b*e*x+a*e)^3/b^2*(e*x+d
)^(1/2)*B*a*d-3/4*e^2/(b*e*x+a*e)^3/b*(e*x+d)^(1/2)*B*d^2+1/8*e^3/(a*e-b*d)/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*
x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*A+5/8*e^3/(a*e-b*d)/b^3/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*
b)^(1/2)*b)*a*B-3/4*e^2/(a*e-b*d)/b^2/((a*e-b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2)*b)*B*d

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 2.12, size = 325, normalized size = 1.56 \[ \frac {e^2\,\mathrm {atan}\left (\frac {\sqrt {b}\,e^2\,\sqrt {d+e\,x}\,\left (A\,b\,e+5\,B\,a\,e-6\,B\,b\,d\right )}{\sqrt {a\,e-b\,d}\,\left (A\,b\,e^3+5\,B\,a\,e^3-6\,B\,b\,d\,e^2\right )}\right )\,\left (A\,b\,e+5\,B\,a\,e-6\,B\,b\,d\right )}{8\,b^{7/2}\,{\left (a\,e-b\,d\right )}^{3/2}}-\frac {\frac {{\left (d+e\,x\right )}^{3/2}\,\left (A\,b\,e^3+5\,B\,a\,e^3-6\,B\,b\,d\,e^2\right )}{3\,b^2}+\frac {\left (a\,e-b\,d\right )\,\sqrt {d+e\,x}\,\left (A\,b\,e^3+5\,B\,a\,e^3-6\,B\,b\,d\,e^2\right )}{8\,b^3}-\frac {{\left (d+e\,x\right )}^{5/2}\,\left (A\,b\,e^3-11\,B\,a\,e^3+10\,B\,b\,d\,e^2\right )}{8\,b\,\left (a\,e-b\,d\right )}}{\left (d+e\,x\right )\,\left (3\,a^2\,b\,e^2-6\,a\,b^2\,d\,e+3\,b^3\,d^2\right )+b^3\,{\left (d+e\,x\right )}^3-\left (3\,b^3\,d-3\,a\,b^2\,e\right )\,{\left (d+e\,x\right )}^2+a^3\,e^3-b^3\,d^3+3\,a\,b^2\,d^2\,e-3\,a^2\,b\,d\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^(3/2))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

(e^2*atan((b^(1/2)*e^2*(d + e*x)^(1/2)*(A*b*e + 5*B*a*e - 6*B*b*d))/((a*e - b*d)^(1/2)*(A*b*e^3 + 5*B*a*e^3 -
6*B*b*d*e^2)))*(A*b*e + 5*B*a*e - 6*B*b*d))/(8*b^(7/2)*(a*e - b*d)^(3/2)) - (((d + e*x)^(3/2)*(A*b*e^3 + 5*B*a
*e^3 - 6*B*b*d*e^2))/(3*b^2) + ((a*e - b*d)*(d + e*x)^(1/2)*(A*b*e^3 + 5*B*a*e^3 - 6*B*b*d*e^2))/(8*b^3) - ((d
+ e*x)^(5/2)*(A*b*e^3 - 11*B*a*e^3 + 10*B*b*d*e^2))/(8*b*(a*e - b*d)))/((d + e*x)*(3*b^3*d^2 + 3*a^2*b*e^2 -
6*a*b^2*d*e) + b^3*(d + e*x)^3 - (3*b^3*d - 3*a*b^2*e)*(d + e*x)^2 + a^3*e^3 - b^3*d^3 + 3*a*b^2*d^2*e - 3*a^2
*b*d*e^2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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